September 2011 - Career Point

(ii) Prove that the harmonic oscillation equation ofthe system is : µ( & x&2 – & x&1) = – k(x 2 – x 1 )where µ =m1mm + m122(iii) What is the oscillation amplitude of (x 2 – x 1 ) ?Sol. The motion of the system of the two bodies can beconveniently described by using the center of massframe of reference. The centre of mass moves in astraight line with constant velocity, due to theconservation of linear momentum. In the centre ofmass frame the two bodies perform simple harmonicoscillations. Denoting the position of the masses m 1and m 2 by x 1 and x 2 , respectively, we can express thedistance between the masses as x 2 – x 1 . The change inthe length of the spring is then x = x 2 – x 1 – a.(i) The forces applied by the spring on the twosystems are :m 1 & x& 1 = kx ...(1)m 2 & x& 2 = – kx ...(2)The signs are used according to the position of themass relative to the spring. Multiplying Eq.(1) by m 2 ,Eq. (2) by m 1 , and subtracting Eq. (1) from Eq. (2)we havem 1 m 2 ( & x&2 – & x&1) = – k(m 1 + m 2 )x ...(3)Since x = x 2 – x 1 – a, we have x& = x& 2 – x&1. And& x& = & x&2 – & x&1. Therefore,m 1 m 2 & x& = –k(m 1 + m 2 )x ...(4)and the solution to this equation is :x(t) = A cos(ω 0 t + φ) ...(5)m1 where ω 0 =+ m2k . Notice that x does notm1m2denote the position of any of the masses. It denotesthe difference between the distance between themasses and the initial state, so that x = (x 2 – x 1 ) – a.The velocity of the centre of mass is given by :v r cm =p10xˆm + m2...(6)(ii) The equation specified in the problem is easilyderived from Eq. (3), which we found in the firstsection. The constant µ is called the "reduced mass"of the system, and is defined as1 1 1 ≡ + ...(7)µ m1m2or in a different form,m1m2µ =...(8)m1+ m2(iii) Taking energy into consideration, we have2 21 ⎛ p0⎞ 1 pE k = m12 ⎜m⎟0= ...(9)⎝ 1 ⎠ 2 m1where E k is the initial kinetic energy. The kineticenergy of the center of mass is :21 p0E k(cm) =...(10)2 m1+ m2and therefore, the total kinetic energy in the centre ofmass frame becomes :m2 2E´k(cm) = E k – E k(cm) =p02m1(m1+ m2)The kinetic energy is proportional to the square ofampitude, E´k = 21 kA2Therefore, A =m2p0km (m + m13. (i) A mass m with kinetic energy E k collides withanother mass M, initially at rest, and sticks to it at themoment of contact. What is the total kinetic energyimmediately after the collision ?(ii) The mass M is used as the weighting surface of aspring-scale whose spring is ideal (a mass lessspring). A body of known mass m is released from acertain height from where it falls to hit M. The twomasses M and m stick together at the moment theytouch, and move together from then on. The oscillationsthey perform reach to height a above the original levelof the scale, and depth b below it (see figure).(a) Find the constant of force of the spring.(b) Find the oscillation frequency.(c) What is the height above the initial level fromwhich the mass m was released ?mmMMambMBefore122)After1Sol. (i) The initial kinetic energy is given by E k = p 2 ,2mand the final value of the kinetic energy is given by1E´k = p´2.2(m + M)Using linear-momentum conservation (p´ = p), we have:mE´k = E km + MXtraEdge for IIT-JEE 16 SEPTEMBER 2011